NETWORK LAYER - Logical Addressing

LOGICAL ADDRESSING IPv4 Addresses
Slide 2Presented by Pankaj Debbarma, Asstt. Prof.
Deptt. of Computer Science & Engineering, TIT
An IPv4 address is a 32-bit address that uniquely and universally defines
the connection of a device (for example, a computer or a router) to the
Internet.

IPv4 ADDRESSES Address Space

Figure 19.1
Dotted-decimal notation and binary notation for an IPv4 address

Example 19.1
Change the following IPv4 addresses from binary notation to dotted-
decimal notation.
a. 10000001 00001011 00001011 11101111
b. 11000001 10000011 00011011 11111111
We replace each group of 8 bits with its equivalent decimal number
and add dots for separation.
a. 129.11.11.239
b. 193.131.27.255

Example 19.2
Change the following IPv4 addresses from dotted-decimal notation to
binary notation.
a. 111.56.45.78
b. 221.34.7.82
We replace each decimal number with its binary equivalent.
a. 01101111 00111000 00101101 01001110
b. 11011101 00100010 00000111 01010010

Example 19.3
Find the error, if any, in the following IPv4 addresses.
a. 111.56.045.78 c. 75.45.301.14
b. 221.34.7.8.20 d. 11100010.23.14.67
a. There must be no leading zero (045).
b. There can be no more than four numbers.
c. Each number needs to be less than or equal to 255.
d. A mixture of binary notation and dotted-decimal notation is not
allowed.

IPv4 ADDRESSES Classful Addressing

Figure 19.2 Finding the classes in binary and dotted-decimal notation

Example 19.3
Find the class of each address.
a. 00000001 00001011 00001011 11101111 c. 14.23.120.8
b. 11000001 10000011 00011011 11111111 d. 252.5.16.111
a. There must be no leading zero (045).
b. There can be no more than four numbers.
c. Each number needs to be less than or equal to 255.
d. A mixture of binary notation and dotted-decimal notation is not
allowed.

Table 19.1 Number of blocks and block size in classful IPv4 addressing

Classful Addressing Transmission Control

Table 19.2 Default masks for classful addressing

Classful Addressing Address Depletion

Classful Addressing Example 19.5
Figure 19.3 shows a block of addresses, in both binary and dotted-decimal
notation, granted to a small business that needs 16 addresses.
We can see that the restrictions are applied to this block. The
addresses are contiguous. The number of addresses is a power of 2
(16 = 24), and the first address is divisible by 16. The first address,
when converted to a decimal number, is 3,440,387,360, which when
divided by 16 results in 215,024,210.

Figure 19.3 A block of 16 addresses granted to a small organization

Classless Addressing Mask

Classless Addressing First Address

Classless Addressing: First Address Example 19.6
A block of addresses is granted to a small organization. We know that one
of the addresses is 205.16.37.39/28. What is the first address in the block?
The binary representation of the given address is 11001101
00010000 00100101 00100111. If we set 32−28 rightmost bits to 0,
we get 11001101 00010000 00100101 00100000 or 205.16.37.32.
This is actually the block shown in Figure 19.3.

Classless Addressing Last Address

Classless Addressing: Last Address Example 19.7
Find the last address for the block in Example 19.6.
The binary representation of the given address is 11001101
00010000 00100101 00100111. If we set 32 − 28 rightmost bits to
1, we get 11001101 00010000 00100101 00101111 or
205.16.37.47.
This is actually the block shown in Figure 19.3.

Classless Addressing Number of Addresses

Classless Addressing: Number of Addresses Example 19.8
Find the number of addresses in Example 19.6.
The value of n is 28, which means that number of addresses is 232−28
or 16.

Another way to find the first address, the last address, and the number of
addresses is to represent the mask as a 32-bit binary (or 8-digit
hexadecimal) number. This is particularly useful when we are writing a
program to find these pieces of information. In Example 19.5 the /28 can
be represented as 11111111 11111111 11111111 11110000 (twenty-
eight 1s and four 0s). Find
a. The first address
b. The last address
c. The number of addresses.

a. The first address can be found by ANDing the given addresses
with the mask. ANDing here is done bit by bit. The result of ANDing
2 bits is 1 if both bits are 1s; the result is 0 otherwise.
Address: 11001101 00010000 00100101 00100111
Mask: 11111111 11111111 11111111 11110000
First Address: 11001101 00010000 00100101 00100000

b. The last address can be found by ORing the given addresses with
the complement of the mask. ORing here is done bit by bit. The
result of ORing 2 bits is 0 if both bits are 0s; the result is 1
otherwise. The complement of a number is found by changing each
1 to 0 and each 0 to 1.
Address: 11001101 00010000 00100101 00100111
Mask: 00000000 00000000 00000000 00001111
First Address: 11001101 00010000 00100101 00101111

c. The number of addresses can be found by complementing the
mask, interpreting it as a decimal number, and adding 1 to it.
Mask complement: 00000000 00000000 00000000 00001111
Number of addresses: 15 + 1 = 16

Figure 19.4 A network configuration for the block 205.16.37.32/28

Classless Addressing Network Address

Figure Hierarchy in a telephone network in Agartala
381 234 2330

Figure 19.6 Two levels of hierarchy in an IPv4 address

Two levels of hierarchy: No Subnetting

Figure 19.7 Configuration and addresses in a subnetted network

Figure 19.8 Three-level hierarchy in an IPv4 address

Classless Addressing: Address Allocation Example 19.10
An ISP is granted a block of addresses starting with 190.100.0.0/16
(65,536 addresses). The ISP needs to distribute these addresses to three
groups of customers as follows:
a. The first group has 64 customers; each needs 256 addresses.
b. The second group has 128 customers; each needs 128 addresses.
c. The third group has 128 customers; each needs 64 addresses.
Design the subblocks and find out how many addresses are still available
after these allocations.

Figure 19.9 shows the situation.
Group 1
For this group, each customer needs 256 addresses. This means that
8(log2 256) bits are needed to define each host. The prefix length is then
32 − 8 = 24. The addresses are
1st Customer: 190.100.0.0/24 190.100.0.255/24
2nd Customer: 190.100.1.0/24 190.100.1.255/24
···
64th Customer: 190.100.63.0/24 190.100.63.255/24
Total = 64 x 256 = 16,384

Group2
1st Customer: 190.100.64.0/25 190.100.64.127/25
2nd Customer: 190.100.64.128/25 190.100.64.255/25
···
128th Customer: 190.100.127.128/25 190.100.127.255/25
Total = 128 x 128 = 16,384

Group 3
1st Customer: 190.100.128.0/26 190.100.128.63/26
2nd Customer: 190.100.128.64/26 190.100.128.127/26
···
64th Customer: 190.100.159.192/26 190.100.159.255/26
Total = 128 x 64 = 8192
Number of granted addresses to the ISP: 65,536
Number of allocated addresses by the ISP: 40,960
Number of available addresses: 24 576

Figure 19.9
An example of address allocation and distribution by an ISP

Table 19.3 Addresses for private networks

Figure 19.10 A NAT implementation

Figure 19.11 Addresses in a NAT

Figure 19.12 NAT address translation

Table 19.4 Five-column translation table

Figure 19.13 An ISP and NAT

NETWORK LAYER - Logical Addressing

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